ÌâÄ¿ÄÚÈÝ

ij»¯Ñ§Ð¡×éÄ£Ä⹤ҵÉú²úÖÆÈ¡HNO3£¬Éè¼ÆÏÂͼËùʾװÖã¬ÆäÖÐaΪһ¸ö¿É³ÖÐø¹ÄÈë¿ÕÆøµÄÏðƤÇò¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Ð´³ö×°ÖÃAÖÐÖ÷Òª·´Ó¦µÄ»¯Ñ§·½³Ìʽ                                   ¡£

£¨2£©ÒÑÖª1molNO2ÓëҺ̬ˮ·´Ó¦Éú³ÉHNO3ÈÜÒººÍNOÆøÌå·Å³öÈÈÁ¿46kJ£¬Ð´³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ                                                ¡£

        ¸Ã·´Ó¦ÊÇÒ»¸ö¿ÉÄæ·´Ó¦£¬ÓûÒªÌá¸ßNO2µÄת»¯ÂÊ£¬¿É²ÉÈ¡µÄ´ëÊ©ÊÇ           ¡£

        A£®½µµÍζȠ    B£®Éý¸ßζȠ           C£®¼õСѹǿ            D£®Ôö´óѹǿ

£¨3£©ÊµÑé½áÊøºó£¬¹Ø±Õֹˮ¼Ðb¡¢c£¬½«×°ÖÃD£¬½þÈë±ùË®ÖУ¬ÏÖÏóÊÇ                     

                                                                           ¡£

£¨4£©×°ÖÃCÖÐŨH2SO4µÄ×÷ÓÃÊÇ                                    ¡£

£¨5£©ÇëÄã°ïÖú¸Ã»¯Ñ§Ð¡×éÉè¼ÆÊµÑéÊÒÖÆÈ¡NH3µÄÁíÒ»·½°¸                                  

                                                           ¡£

£¨6£©¸ÉÔï¹ÜÖеļîʯ»ÒÓÃÓÚ¸ÉÔïNH3£¬Ä³Í¬Ñ§Ë¼¿¼ÊÇ·ñ¿ÉÓÃÎÞË®ÂÈ»¯¸Æ´úÌæ¼îʯ»Ò£¬²¢Éè¼ÆÓÒͼËùʾװÖã¨ÒÇÆ÷¹Ì¶¨×°ÖÃÊ¡ÂÔδ»­£©½øÐÐÑéÖ¤¡£ÊµÑé²½ÖèÈçÏ£º

   ¢ÙÓÃÉÕÆ¿ÊÕ¼¯Âú¸ÉÔïµÄ°±Æø£¬Á¢¼´ÈûÉÏÈçͼµÄʾµÄÏð½ºÈû¡£

¢ÚÕýÁ¢ÉÕÆ¿£¬Ê¹ÎÞË®ÂÈ»¯¸Æ¹ÌÌ廬ÈëÉÕÆ¿µ×²¿£¬Ò¡¶¯£¬¿É¹Û²ìµ½

µÄÏÖÏóÊÇ              £¬

ÓÉ´Ë£¬¸ÃͬѧµÃ³ö½áÂÛ£º²»ÄÜÓÃCaCl2´úÌæÊ¯»Ò¡£

(1)2NH4Cl+Ca(OH)2 ¡÷ 2NH3¡ü+CaCl2+2H2O

(2)3NO2(g)+H2O(l)£½2HNO3(aq)+NO(g)  ¡÷H=-138kJ?mol-1

AD

(3)ÑÕÉ«±ädz

(4)ÎüÊÕ¶àÓàµÄNH3

(5)·½°¸I£º¼ÓÈÈŨ°±Ë®ÖÆÈ¡NH3¡£  ·½°¸¢ò£ºÏòNaOH»òCaO¹ÌÌåÉϵμÓŨ°±Ë®ÖÆÈ¡NH3¡£(ÆäËûºÏÀí´ð°¸Ò²¸ø·Ö)

(6)ÆøÇòÅòÕÍ

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø